Question

You are traveling in a convertible with the top down. The car is moving at a constant velocity 22.5 m/s, due east along flat ground. You throw a tomato straight upward at a speed of 10.6 m/s. How far has the car moved when you get a chance to catch the tomato?

Answer 1

The car moved a distance
`x=48.6m` .

Step-by-step explanation:

First we need to know: How much time will the tomato spend in the air?

From Kinematics:
`v_y(t)=v_0_y+at`

where
`v_0_y=10.6(m)/(s)` and
`a=-g=-9.8(m)/(s^2)` is gravity's acceleration.

`v_y(t)=v_0_y-gt`

When the tomato touches the car again,
`v_y=-v_0_y=-10.6(m)/(s)`

Then, we have:

`t=(v_y-v_o_y)/(-g)` ⇒
`t=(v_y-v_o_y)/(-g)=2.16s`

Also from Kinematics we have:
`x(t)=v_xt`

Which is very simple because we can take initial position 0 and there's no acceleration in the x direction. And
`v_x=22.5(m)/(s)`

So, taking
`t=2.16s`

`x=48.6m`